Two gases, P and Q, are contained in separate vessels. The volume of the vessel containing gas P is $1 \, dm^3$ and the volume of the vessel containing gas Q is $2 \, dm^3$. The temperature and pressure of both gases are the same. The mass of both gas samples are also the same. We need to find the ratio of the molar masses of P and Q.

1. Problem Description

Two gases, P and Q, are contained in separate vessels. The volume of the vessel containing gas P is $1 \, dm^3$ and the volume of the vessel containing gas Q is $2 \, dm^3$. The temperature and pressure of both gases are the same. The mass of both gas samples are also the same. We need to find the ratio of the molar masses of P and Q.

2. Solution Steps

Let $V_P$ be the volume of gas P, and $V_Q$ be the volume of gas Q.

$V_P = 1 \, dm^3$

$V_Q = 2 \, dm^3$

Let $T$ be the temperature and $P$ be the pressure. Since both gases are at the same temperature and pressure, $T_P = T_Q = T$ and $P_P = P_Q = P$.

Let $m_P$ be the mass of gas P and $m_Q$ be the mass of gas Q.

Given that $m_P = m_Q = m$.

We can use the ideal gas law:

$PV = nRT$

where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the ideal gas constant, and $T$ is the temperature.

For gas P, $PV_P = n_P RT$

For gas Q, $PV_Q = n_Q RT$

Dividing the equation for gas P by the equation for gas Q, we get:

$\frac{PV_P}{PV_Q} = \frac{n_P RT}{n_Q RT}$

$\frac{V_P}{V_Q} = \frac{n_P}{n_Q}$

$\frac{1}{2} = \frac{n_P}{n_Q}$

$n_Q = 2n_P$

The number of moles is related to the mass and molar mass by the formula:

$n = \frac{m}{M}$

where $m$ is the mass and $M$ is the molar mass.

For gas P, $n_P = \frac{m_P}{M_P} = \frac{m}{M_P}$

For gas Q, $n_Q = \frac{m_Q}{M_Q} = \frac{m}{M_Q}$

Substituting these into the previous equation $n_Q = 2n_P$:

$\frac{m}{M_Q} = 2 \frac{m}{M_P}$

$\frac{1}{M_Q} = \frac{2}{M_P}$

$M_P = 2M_Q$

$\frac{M_P}{M_Q} = 2$

Therefore, the ratio of the molar masses of P and Q is $2:1$.

3. Final Answer

2:1